How many passwords are possible in this scenario?

Question

A password string consists of one or more of the 26 characters A..Z and can be of any length from $1$ to $8$ characters. How many password are possible in this scenario?

• As my calculation, it should be $2^{26}\times8!$, is it correct?

Answer 1

• for one 1 char passwords, one can only put $26$ different characters
• for one 2 char passwords, one can only put $26^2$ different characters
• for one 3 char passwords, one can only put $26^3$ different characters
• ...
• for one 8 char passwords, one can only put $26^8$ different characters

In short, think each box can contain how many letters from the given alphabet and multiply the values in the boxes. Finally, sum them.

$$\sum_{i=1}^8 26^i = \frac{26^9-1}{26-1}-1$$

The name of the formula is Sum of Consecutive powers of a number;

$$\sum_{i=0}^n a^i = \dfrac{a^{n+1} -1}{a-1}$$

and $-1$ since $i=0$ is not a case here.

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Update : a more general case;

Let assume that you want to build a web application and want to determine the security level of the passwords. Here a calculator;

• Assume that the alphabet has $l$ letters,
• $m$ numerals, and
• $p$ non-alphanumeric.

If we set the passwords to require at least;

• $n_m$ numerals
• $n_p$ non-alphanumeric,with
• passords's length to $$min \geq n_k+n_a \geq 6,\text{and}$$
• $max > min$, then

What is the password space, $\mathcal{S}$?

$$ \mathcal{S} = m^{n_m} + p^{n_p} + \sum_{i=1}^{max - ({n_m} + {n_p})} (l+m+p)^i$$

Under these assumptions, let

• $l = 26$

• $m = 10$

• $p = 20$

• $min = 6$

• $max=20$

• $n_m=2$

• $n_p =1$, then we have;

  $$ \mathcal{S}_{6:20} = 10^2 + 20 + \sum_{i=1}^{17} (56)^i = 533361663473057950558105648760 $$

$ \mathcal{S}_{6:20}$ has 99 digits in binary form.

if max is;

• $6 \text{ then } S = 178928$
• $7 \text{ then } S = 10013424$
• $8 \text{ then } S = 10013424$
• $9 \text{ then } S = 560745200$
• $10\text{ then } S = 31401724656$ is 35-bit.

WolframAlpha calculator;

m^{n_m} + p^{n_p} + sum (l+m+p)^i, i=1 to (max - (n_m + n_p))

Answer 2

No, it is incorrect. As I assume this is homework. I won't supply a full answer.

$2^{26}$ suggests 26 binary choices. But that is not the case.
$8!$ would suggest an arbitrary ordering of 8 characters.

What you have is the 8 different lengths, you can sum the number of combinations for each length.

Note letters can repeat themselves; you can select letters independently.